3.
m₁=3 kg at (0,0) m₂= 4 kg at (2,1) and m₃=8 kg at (1,2)
(x_cm,y_cm) = ¹/_MSm_i*(x_i,y_i) = [1/(3+4+8)]*[3* (0,0) + 4*(2,1) + 8*(1,2)] = ¹/₁₅(8,20) = ( ⁸/₁₅,²⁰/₁₅)
4.
Each rod can be treated as a mass M (3M) concentrated in the center of mass for this rod.  The center of mass of an uniform rod is always at the center of length (symmetry).
In the coordinate system shown (x-horizontal) left rod has its center of mass at (-^L/₂,^L/₂), right rod has its center of mass at (^L/₂,^L/₂), and the upper rod has its center of mass at (0,L) (from symmetry).
(x_cm,y_cm) = ¹/_MSm_i*(x_i,y_i) = [¹/_(M+M+3M)]*[M* (-^L/₂,^L/₂) + M*(^L/₂,^L/₂) + 3M*(0,L)] = ¹/_5M*(0,4ML) = (0,⁴/₅L)
30.

a)    K_i = 2100*(41*1000/3600)²     K_f = 2100*(51*1000/3600)²   DK = K_f -K_i
b)  in coordinate system    x to East and y to North:
 p = mv   so p_i = 2100*(0i+41*1000/3600j) p_f = 2100*(51*1000/3600i+0j)   Dp = p_f -p_i
34.
System shown was initially at rest.  If no external forces are acting on system (friction is negligible) then the linear momentum is conserved.  Initial momentum of system is zero so is the final.
 0.15*(+v_s) + 200*(-v_m) = 0
 
 

35.
If center of mass remains at rest >> total linear momentum = 0
P = 1kg*(+1.7) + 3kg*(-v) = 0
 

46.
If no external forces are acting on system (explosion is internal event)) then the linear momentum is conserved  (P_i = P_f).   The directions of v_A and  v_B is guessed.  If I am right the result should be positive, if I am not - the result will be negative.
   8*v = 4*v_A + 4*v_B
and we know  Kinetic Energy after explosion.  It is ¹/₂*8*v² +16 =32 J
32 = ¹/₂*4*v_A² + ¹/₂*4*v_B²   Two equations and two unknown.  Solve for v_A and v_B
47.
m₁ = ^M/₆  and m₂ = ^M/₁₂

a)      m₁+ m₂ = ^M/₄     from conservation of linear momentum:    ^M/₄*v + M*[-(v_eff -v)] = 0
solving for v = ⁴/₅ v_eff
b)   from conservation of linear momentum:  m₁*v +(M+m₂)*[-(v_eff -v)] = 0
 ^M/₆*v +(M+^M/₁₂)*[-(v_eff -v)] = 0     so      v = ¹³/₁₅*v_eff
and velocity of iceboat is:       v_eff - ¹³/₁₅*v_eff = ²/₁₅*v_eff = 0.13v_eff

now for mass m₂
     from conservation of linear momentum:  m₂*V +M*[-(v_eff -V)] = (M+m₂)(-²/₁₅*v_eff )
^M/₁₂*V +M*[-(v_eff -V)] = (M+^M/₁₂)*(-²/₁₅*v_eff ) so   V = 0.79 v_eff  and the speed of boat is  v_eff  - 0.79 v_eff  = 0.21 v_eff

c) ?

54.
  It is 540kg/min mass that has to have momentum changed.  Initial is 0 and final is 540kg*3.2m/s in 60 seconds. S F = dP/dt = (540kg*3.2m/s) /60s = 28.8 N